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MATH 265
(MATHEMATICAL METHODS)
Rev. Dr. W. Obeng Denteh
November 20, 2015
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 1 / 20
Outline
1 Partial Differentiation Of Function Several Variables
2 Differentiation of Implicit Functions
3 Theorem and Application
4 Jacobians
5 Differentiation of a Vector Function of several variables
6 The Tangent Vector
7 Curvilinear Coordinates
8 Multiple Surface and Volume Integral
9 Gradient, Divergence and Curls.
10 The theorems of Green, Gauss and Stokes.
11 Applications to physical and Geometrical problems.
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 1 / 20
Partial Differentiation Of Function Several
Variables
The Volume V of a Cylinder of radius r and height h is given by:
V = πr 2h
i.e V depends on two quantities r and h.The volume V increases if r is kept constant as h is increased.Considering the derivative of V with respect to h when r is keptconstant is written as
∂V
∂h= πr 2
For vice versa,the derivative of V with respect to r when r is keptconstant is written as
∂V
∂r= 2πrh
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 2 / 20
Therefore,∂V
∂h
and∂V
∂r
are called the Partial Derivatives of h and r respectively. V is thedependent variable whereas r and h are the independent variables.V is therefore a function of (x , y) which can be written asV = (x , y) The following can be defined from V = (x , y)∂v∂x
∂v∂y
∂2v∂x2
∂2v∂x2
∂2v∂y .∂x
∂2v∂x .∂y
And also, ∂2v∂y .∂x
∂2v∂x .∂y
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 3 / 20
Examples1 Given u = x2 + xy + y 2, find du
dxand du
dy.
Solution
∂u
∂x= 2x + y
∂u
∂y= x + 2y
2 Find the partial derivatives of the function, z = x3 + y 3 − 2x2ywith respect to the variables x and y.Solution
∂z
∂x= 3x2 − 4xy
∂z
∂y= 3y 2 − 2x2
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 4 / 20
Exercise 1
Find ∂z∂x, ∂z∂y
of the following functions:
1 z = (2x − y)(x + 3y)
2 z = 2x−yx+y
3 f (x , y , z) = xsin(y + 3z).
4 z = tan(3x + 4y)
5 z = sin(3x+2y)xy
6 Given V = ln(x2 + y 2), prove that ∂2v∂x2
+ ∂2v∂y2 = 0
7 find the first and second partial derivatives of the function:z = x3 + y 3 − 2x2y
8 If V = f (x2 + y 2), Show that x ∂V∂y− y ∂V
∂x= 0
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 5 / 20
Differenciation Of Implicit Functions
If y = x2 − 4x + 2, y is completely defined in terms of x , and y iscalled an explicit function of x .When the relationship between x and y is more involved, it may notbe possible to separate y completely on the left-hand side, eg.xy + siny = 2. In such a case as this, y is called an implicit functionof x because a relationship of the form y = f (x) is implied in thegiven equation.
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 6 / 20
Example
If x2 + y 2 − 2− 6y + 5 = 0, find dydx
and d2ydx2
at x = 3, y = 2.Solution.
2x + 2ydy
dx− 2− 6
dy
dx= 0
(2y − 6)dy
dx= 2− 2x
dy
dx=
2− 2x
2y − 6=
1− x
y − 3
∴ at(3, 2),dy
dx=
1− (3)
(2)− 3= 2
Thend2y
dx2=
d
dx
{1− x
y − 3
}=
(y − 3)(−1)− (1− x)dydx
(y − 3)2
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 7 / 20
Example Cont’.
=(3− y)− (1− x)dy
dx
(y − 3)2
at(3, 2)d2y
dx2=
(3− 2)− (1− 3)2
(y − 3)2=
1− (−4)
1= 5
∴ At (3, 2), dydx
= 2 and d2ydx2
= 5
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 8 / 20
Exercise 21 Find df
dxand df
dyof the following:
(i) f (x , y) = logy x(ii) f (x , y) =
∫ yx g(t)dt
(iii) f (x , y) = 2x2 − 3y − 4(iv) f (x , y) = sin2(x − 3x)
2 Find fy , fx and fz of the following:
(i) f (x , y , z) = (x2 + y2 + z2)−12
(ii) f (x , y , z) = e(x2+y2+z2)
3 Find the value of dzdx
at the point (1,-1,-3) if the equationxz + ylnx − z2 + 4 = 0 defines x as a function of the twoindependent variables y and z and the partial derivatives exists.
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 9 / 20
Theorem and Application
Clairaut’s TheoremThis theorem states that: Given f (x , y) such that its partialderivatives fx , fy , fxy and fyx are defined throughout an open regioncontaining a point (a, b) and are all continuous at (a, b) then;
fxy (a, b) = fyx(a, b)
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 10 / 20
Example
Verify Clairaut Theorem for f (x , y) = yln(4x + y 2)Solution
fx =4y
4x + y 2
fxy = (4x + y 2)4− 4y(2y)
=16x + 4y 2 − 8y 2
(4x + y 2)2
fyx = 2y 2(−1)(4x + y 2)−24 +4
4x + y 2
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 11 / 20
Application/Interpretation of Partial Differential
Equation
Geometrical Interpretation Of Partial Derivatives
Let z = f (x , y) be a surface.z = f (x , y0) is a curve (where y0 is a constant). This curvez = f (x , y) is obtained by the intersection of the surface z = f (x , y)with the plane y = y0. df
dxis the slope of the tangent line to the curve
z = f (x , y0).Example Given z = f (x , y) = x2
y3
(a) Find the slope of the tangent lines at (2,5)for
(i) y fixed and (ii) x fixed
(b) Determine whether z = f (x , y) is increasing or decreasing at(2,5) for
(i) y fixed and (ii) x fixed
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 12 / 20
Solution(ai) When y is fixed means differentiating f w.r.t x whiles y is kept
constant. Thus:df (x ,y)
dx= 2x
y3 = df (2,5)dx
= 2(2)53
= 4125
Hence the slope of the tangent line at (2,5) for fixed y is 4125
(aii) When x is fixed means differentiating f w.r.t y whiles x is keptconstant. Thus:df (x ,y)
dy= −3x2y−4 = −3x2
y4 = df (2,5)dy
= −3(2)254
= −12625
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 13 / 20
(bi) fx = 4125
> 0⇒ f (x , y)df (2,5)dy
= −12625
< 0 is increasing
(bii) Decreasing
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 14 / 20
Exercise 31 Given f (xy) = ex2y . Verify the Clairaut’s theorem
2 Find the slopes of the traces to z = 10− 4x2 − y 2 at the point(1,2).
3 Find the vector equations of the tangent lines to the traces toz = 10− 4x2 − y 2 at the point (1,2).
4 Verify Clairaut’s Theorem for the function f (x , y) = xe−x2y2
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 15 / 20
Jacobian
Implicit Differentiation Using JacobiansConsider the system
F (u, v , x , y) = 0
G (u, v , x , y) = 0
Where u = u(x , y) and v = v(x , y).To find ......... for the above system we proceed as follows:Find expressions for ... and... in a step by step manner and deduceexpressions for the others.Using implicit function differentiation for each of the equations in theabove system we have;
d
dx[F (u, v , x , y)] = Fx + Fuux + Fvvx = 0
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 16 / 20
Jacobian Cont’d
dx[G (u, v , x , y)] = Gx + Guux + Gvvx = 0
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 17 / 20
Differentiation of a Vector function of several
variables
Let (a, b) be x − y coordinates of a point on the surface z = f (x , y).Then the equation of the surface as a vector function is
ρ(x , y) =< x , y , z >=< x , y , f (x , y) >
The tangent vector for traces with fixed y is given asρx(x , y) =< 1, 0, fx(x , y) >. The tangent vector for traces with fixedx is given as ρy (x , y) =< 0, 1, fy (x , y) >.NB:Trace- The point at which a line or the curve in which a surfaceintersects a coordinate plane.
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 18 / 20
Cont’The equation for the tangent line to traces with fixed ρ is then
ρ(t) =< a, b, f (a, b) > +t < 1, 0, fx(a, b) >
Examples Find the vector equation of the tangent lines
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 19 / 20
The Tangent Vector
ExampleFind a tangent vector at the point where t = 3 on the vectorf (t) = e2t i + (t2 − 1)j + lntk
Solution
f ′(t) = 2e2t i + 2tj +1
tk
f ′(t)|t=3 = 2e6i + 6j +1
3k
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 20 / 20
Gradient, Divergence and Curls.
GradientThe gradient of a scalar point function f = f (x , y , z) is defined b:
Gradf = ∇f =
(i
d
dx+ j
d
dy+ k
d
dz
)= i
df
dx+ j
df
dy+ k
df
dz
Gradf , ∇f is a vector quantity.
df =df
dxdx +
df
dydy + k
df
dzdz
∇f =
(idf
dx+ j
df
dy+ k
df
dz
)· (idx + jdy + dzk)
= ∇f
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 21 / 20
= |∇f |cos θ
θ is the angle between ∇f and dr .The value of df is greatest when θ = 0, or cos θ = 1.This gives the gradient of f to ∇f .
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 22 / 20
Example
If φ = 3x2yy − y 3z2 , find ∇φ at the point (1,-2,-1).Solution.
∇f =
(idφ
dx+ j
dφ
dy+ k
dφ
dz
)= 6xy i + (3x2 − 3y 2z2)j + (−2y 3z)k
φ|(1,−2,1) = −12i− 9j− 16k
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 23 / 20
DivergenceThe divergence of a vector point function F is denoted by div Fwhere,
divF = ∇·F =
(i
d
dx+ j
d
dy+ k
d
dz
)· (F1i + F2j + F3k)
Clearly, div is a scalar. If div = 0, F is called Solenoidal.ExampleEvaluate div(3x2i + 5xy 2j + xyz3k) at the point (1,2,3).Solution
div(3x2i+5xy 2j+xyz3k) =
(i∂
∂x+ j
∂
∂y+ k
∂
∂z
)· (3x2i+5xy 2j+xyz3k)
= 6x + 10xy + 3xyz2
div(3x2i + 5xy 2j + xyz3k)|(1,2,3) = 6 + 20 + 54 = 80Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 24 / 20
CurlThe Curl of a vector point function F is defined as: curlF = ∇× F .If F1i + F2j + F3k, then:
curlF = ∇× F
is given as (∂i
∂x+∂j
∂y+∂k
∂z
)(F1i + F2j + F3k)
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 25 / 20
Exercise1 Find the unit normal to the surface x2 + y 2 − z = 1 at P(1,1,1).
2 Let v = (x + y)i+ (y − 2z)j+ (x +λ)k. Find λ if v is solenoidal.
3 If v = x i+y j+zk√x2+x2+z2
, find λ· v
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 26 / 20
The Theorems Of Green, Gauss and Stokes.
Green’s TheoremLet C be a positively oriented, piecewise smooth, simple, closed curveand let D be the region enclose by the curve. If P and Q havecontinuous first order partial derivatives on D then,∫
C
Pdx + Qdy =
∫ ∫D
(∂Q
∂x− ∂P
∂y
)dA
There are some alternate notations to be acknowledged beforeworking some examples.When working with a line integral in which the path satisfies thecondition of Green’s Theorem, we will often denote the line integralas, ∫
Pdx + Qdy or
∫Pdx + Qdy
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 27 / 20
Green’s Theorem Cont’Both of these notations do assume that C satisfies the conditions ofGreen’s Theorem.Also, sometimes the curve C is not thought of as a separate curvebut instead as the boundary of some region D and in these cases Cmay be denoted ad ∂D.
Example
Use Green’s Theorem to evaluate∫C
xydx + x2y 3dy where C is thetriangle with vertices (0,0), (1,0),(1,2).
SolutionThe following inequalities will define the region enclosed.
0 ≤ x ≤ 1 0 ≤ y ≤ 2x
We can identify P and ! from the line integral. Here they are.
P = xy Q = x2y 3Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 28 / 20
So the Green’s Theorem of line integral becomes,∫C
xydx + x2y 3dy =
∫ ∫D
(2xy 3 − x)dA =
∫ 1
0
∫ 2x
0
(2xy 3 − x)dydx
=
∫ 1
0
(1
2xy 4 − xy
)|2x0 dx
Rev. Dr. W. Obeng Denteh (KNUST) MATH 265(MATHEMATICAL METHODS) November 20, 2015 29 / 20